Ordinary Differential Equations

Often in Physics, Chemistry, Biology, Geometry, etc there arise equations that relate a function with its derivative, or successive derivatives.

Definition - Ordinary differential equation. An ordinary differential equation (O.D.E.) is a equation that relates an independent variable $x$, a function $y(x)$ that depends on $x$, and the successive derivatives of $y$, $y’,y’’,\ldots,y^{(n)}$; it can be written as \[F(x, y, y’, y’’,\ldots, y^{(n)})=0.\] The order of a differential equation is the greatest order of the derivatives in the equation.

Thus, for instance, the equation $y’’‘+sen(x)y’=2x$ is a differential equation of order 3.

Deducing a differential equation

To deduce a differential equation that explains a natural phenomenon is essential to understand what a derivative is and how to interpret it.

Example. Newton’s law of cooling states

“The rate of change of the temperature of a body in a surrounding medium is proportional to the difference between the temperature of the body $T$ and the temperature of the medium $T_a$.”

The rate of change of the temperature is the derivative of temperature with respect to time $dT/dt$. Thus, Newton’s law of cooling can be explained by the differential equation \[\frac{dT}{dt}=k(T-T_a),\] where $k$ is a proportionality constant.

Solution of an ordinary differential equation

Definition - Solution of an ordinary differential equation. Given an ordinary differential equation $F(x,y,y’,y’’,\ldots,y^{(n})=0$, the function $y=f(x)$ is a solution of the ordinary differential equation if it satisfies the equation, that is, if \[F(x,f(x), f’(x), f’‘(x),\ldots, f^{(n}(x))=0.\] The graph of a solution of the ordinary differential equation is known as integral curve.

Solving an ordinary differential equations consists on finding all its solutions in a given domain. For that integral calculus is required.

The same manner than the indefinite integral is a family of antiderivatives, that differ in a constant term, after integrating an ordinary differential equation we get a family of solutions that differ in a constant. We can get particular solutions giving values to this constant.

General solution of an ordinary differential equation

Definition - General solution of an ordinary differential equation. Given an ordinary differential equation $F(x,y,y’,y’’,\ldots,y^{(n})=0$ of order $n$, the general solution of the differential equation is a family of functions \[y =f (x,C_1,\ldots,C_n),\] depending on $n$ constants, such that for any value of $C_1,\ldots,C_n$ we get a solution of the differential equation.

For every value of the constant we get particular solution of the differential equation. Thus, when a differential equation can be solved, it has infinite solutions.

Geometrically, the general solution of a differential equation corresponds to a family of integral curves of the differential equation.

Often, it is common to impose conditions to the solutions of a differential equation to reduce the number of solutions. In many cases, these conditions allow to determine the values of the constants in the general solution to get a particular solution.

First order differential equations

In this chapter we are going to study first order differential equations \[F(x,y,y’)=0.\]

The general solution of a first order differential equation is \[y = f (x,C),\] so to get a particular solution from the general one, it is enough to set the value of the constant $C$, and for that we only need to impose one initial condition.

Definition - Initial value problem. The group consisting of a first order differential equation and an initial condition is known as initial value problem:

Solving an initial value problem consists in finding a solution of the first order differential equation that satisfies the initial condition.

Example. Recall the first order differential equation of the Newton’s law of cooling, \[\frac{dT}{dt}=k(T-T_a),\] where $T$ is the temperature of the body and $T_a$ is the temperature of the surrounding medium.

It is easy to check that the general solution of this equation is \[T(t) = Ce^{kt}+T_a.\]

If we impose the initial condition that the temperature of the body at the initial instant is $5$ ºC, that is, $T(0)=5$, we have \[T(0) = Ce^{k\cdot0}+T_a = C+T_a = 5,\] from where we get $C=5-T_a$, and this give us the particular solution \[T(t) = (5-T_a)e^{kt}+T_a.\]

Integral curve of an initial value problem

Example. If we assume in the previous example that the temperature of the surrounding medium is $T_a=0$ ºC and the cooling constant of the body is $k=1$, the general solution of the differential equation is \[T(t)=Ce^t,\] that is a family of integral curves. Among all of them, only the one that passes through the point $(0,5)$ corresponds to the particular solution of the previous initial value problem.

Area of a positive function

Existence and uniqueness of solutions

Theorem - Existence and uniqueness of solutions of a first order ODE. Given an initial value problem

if $F(x,y(x))$ is a function continuous on an open interval around the point $(x_0,y_0)$, then a solution of the intial value problem exists. If, in addition, $\frac{\partial F}{\partial y}$ is continuous in an open interval around $(x_0,y_0)$, the solution is unique.

Although this theorem guarantees the existence and uniqueness of a solution of a first order differential equation, it does not provide a method to compute it. In fact, there is not a general method to solve first order differential equations, but we will see how to solve some types:

Separable differential equations

Definition - Separable differential equation. A separable differential equation is a first order differential equation that can be written as \[y’g(y)=f(x),\] or what is the same, \[g(y)dy=f(x)dx,\] so the different variables are on different sides of the equality (the variables are separated).

The general solution for a separable differential equation comes after integrating both sides of the equation \[\int g(y)\,dy = \int f(x)\,dx+C.\]

Example. The differential equation of the Newton’s law of cooling \[\frac{dT}{dt}=k(T-T_a),\] is a separable differential equation since it can be written as \[\frac{1}{T-T_a}dT=k\,dt.\]

Integrating both sides of the equation we have \[\int \frac{1}{T-T_a}\,dT=\int k\,dt\Leftrightarrow \log(T-T_a)=kt+C,\] and solving for $T$ we get the general solution of the equation \[T(t)=e^{kt+C}+T_a=e^Ce^{kt}+T_a=Ce^{kt}+T_a,\] rewriting $C=e^C$ as an arbitrary constant.

Homogeneous differential equations

Definition - Homogeneous function. A function $f(x,y)$ is homogeneous of degree $n$, if it satisfies \[f(kx,ky)= k^nf(x,y),\] for any value $k\in \mathbb{R}$.

In particular, a homogeneous function of degree $0$ always satisfies \[f(kx,ky)=f(x,y).\]

Setting $k=1/x$ we have \[f(x,y)=f\left(\frac{1}{x}x,\frac{1}{x}y\right)=f\left(1,\frac{y}{x}\right)=g\left(\frac{y}{x}\right).\] This way, a homogeneous function of degree $0$ always can be written as a function of $u=y/x$: \[f(x,y)=g\left(\frac{y}{x}\right)=g(u).\]

Definition - Homogeneous differential equation A homogeneous differential equation is a first order differential equation that can be written as \[y’=f(x,y),\] where $f(x,y)$ is a homogeneous function of degree $0$.

We can solve a homogeneous differential equation by making the substitution \[u=\frac{y}{x}\Leftrightarrow y=ux,\] so the equation becomes \[u’x+u=f(u),\] that is a separable differential equation.

Once solved the separable differential equation, the substitution must be undone.

Example. Let us consider the following differential equation \[4x-3y+y’(2y-3x)=0.\]

Rewriting the equation in this way \[y’=\frac{3y-4x}{2y-3x}\] we can easily check that it is a homogeneous differential equation.

To solve this equation we have to do the substitution $y=ux$, and we get \[u’x+u=\frac{3ux-4x}{2ux-3x}=\frac{3u-4}{2u-3}\] that is a separable differential equation.

Separating the variables we have \[u’x=\frac{3u-4}{2u-3}-u=\frac{-2u^2+6u-4}{2u-3}\Leftrightarrow \frac{2u-3}{-2u^2+6u-4}\,du=\frac{1}{x}\,dx.\]

Now, integrating both sides of the equation we have

then, applying the exponential function to both sides and simplifying we get the general solution \[u^2-3u+2=e^{-2\log|x|-2C}=\frac{e^{-2C}}{e^{\log|x|^2}}=\frac{C}{x^2},\] rewriting the constant $K=e^{-2C}$.

Finally, undoing the initial substitution $u=y/x$, we arrive at the general solution of the homogeneous differential equation \[\left(\frac{y}{x}\right)^2-3\frac{y}{x}+2=\frac{K}{x^2}\Leftrightarrow y^2-3xy+2x^2=K.\]

Linear differential equations

Linear differential equations

Definition - Linear differential equation A linear differential equation is a first order differential equation that can be written as \[y’+g(x)y = h(x).\]

To solve a linear differential equation we try to write the left side of the equation as the derivative of a product. For that we multiply both sides by the function $f(x)$, such that \[f’(x)=g(x)f(x).\] Thus, we get

Integrating both sides of the previous equation we get the solution \[yf(x)=\int h(x)f(x)\,dx+C.\] On the other hand, the unique function that satisfies $f’(x)=g(x)f(x)$ is \[f(x)=e^{\int g(x)\,dx},\] so, substituting this function in the previous solution we arrive at the solution of the linear differential equation \[ye^{\int g(x)\,dx}=\int h(x) e^{\int g(x)\,dx}\,dx+C,\] or what is the same \[y=e^{-\int g(x)\,dx}\left(\int h(x)e^{\int g(x)\,dx}\,dx+C\right). \]

Example. If in the differential equation of the Newton’s law of cooling the temperature of the surrounding medium is a function of time $T_a(t)$, then the differential equation \[\frac{dT}{dt}=k(T-T_a(t)),\] is a linear differential equation since it can be written as \[T’-kT=-kT_a(t),\] where the independent term is $-kT_a(t)$ and the coefficient of $T$ is $-k$.

Substituting in the formula of the general solution of a linear differential equation we have \[y=e^{-\int -k\,dt}\left(\int -kT_a(t)e^{\int -k\,dt}\,dt+C\right)= e^{kt}\left(-\int kT_a(t)e^{-kt}\,dt+C\right).\]

In the particular case that $T_a(t)=t$, and the proportionality constant $k=1$, the general solution of the linear differential equation is \[y=e^{t}\left(-\int te^{-kt}\,dt+C\right)=e^t(e^{-t}(t+1)+C)=Ce^t+t+1.\] If, in addition, we know that the temperature of the body at time $t=0$ is $5$ ºC, that is, we have the initial condition $T(0)=5$, then we can compute the value of the constant $C$, \[y(0)=Ce^0+0+1=5 \Leftrightarrow C+1=5 \Leftrightarrow C=4,\] and we get the particular solution \[y(t)=4e^t+t+1.\]